Modulus and Conjugate of a Complex Number

The modulus of a complex number gives the distance of the complex number from the origin in the Argand plane, whereas the conjugate of a complex number gives the reflection of the complex number about the real axis in the Argand plane. In this section, we will discuss the modulus and conjugate of a complex number, along with a few solved examples.

Conjugate of a Complex Number

The conjugate of a complex number z = x + iy is denoted by

The geometrical representation of the complex number is shown in the figure given below:

Properties of the Conjugate of a Complex Number

Below are some properties of the conjugate of complex numbers, along with their proof.

(1)

Proof: Let z₁ = a + ib and z₂ = c + id

Then,

= a ± c – i(b ± d)

= a – ib ± c ± id

= a – ib ± (c – id)

(2)

Modulus of a Complex Number

The modulus of the complex number is the distance of the point on the Argand plane representing the complex number z from the origin.

Let P is the point that denotes the complex number, z = x + iy.

Then, OP = |z| = √(x² + y² ).

Properties of Modulus of Complex Number

Below are a few important properties of the modulus of a complex number and their proofs.

(i) |z₁ z₂| = |z₁||z₂|

Proof: let z₁= a + ib and z₂ = c + id

Then, |z₁ z₂| = |(a + ib)(c + id)|

⇒ |ac + iad + ibc + i²bd|

⇒ |ac + iad + ibc – bd|

⇒ |ac – bd + i(ad + bc)|

⇒ (ac – bd)²+ (ad+bc)²

⇒ (ac)²+ (bd)² – 2abcd + (ad)² + (bc)² + 2abcd

⇒ a² c² + b² d² + a² d²+ b² c²

⇒ a² c² + b² c² + b² d² + a² d²

⇒ (a² + b²)c² + (b² + a²)d²

⇒ (a² + b²) (c² + d²)

⇒ |z₁||z₂|.

(ii) |z₁ / z₂ | = (|z₁|) / (|z₂|).

Proof: |z₁/z₂ | = |z₁ . 1/z₂ |

Using the multiplicative property of modulus, we have

⇒ |z₁| |1/z₂|

⇒ |z₁| 1/(|z₂|)

⇒ (|z₁ |) / (|z₂|).

Also, read

• Complex numbers
• Complex numbers solved examples

Examples on Modulus and Conjugate of a Complex Number

Example 1: Find the conjugate of the complex number z = (1 + 2i)/(1 – 2i).

Solution: z = (1 + 2i)/(1 – 2i)

Rationalising given the complex number, we have

⇒ z = ((1 + 2i)/(1 – 2i) )× (1 + 2i)/(1 + 2i)

⇒ z = (1 + 2i)²/(1² – (2i)²)

⇒ z = (1 + 4i² + 4i)/(1 + 4)

⇒ z = (1 – 4 + 4i)/(1 + 4)

⇒ z = (-3 + 4i)/5

Example 2: Find the modulus of the complex number z = (3 – 2i)/2i

Solution: z = (3 – 2i)/2i

⇒ z = (3 )/2i – 2i/2i

⇒ z = 3/2i – 1

⇒ z = 3i/(2i² ) – 1

⇒ z = (-3i/2) – 1

Example 3: If z + |z| = 1 + 4i, then find the value of |z|.

Solution: Let z = x + iy

⇒ z + |z| = 1 + 4i

⇒ x + iy + |x + iy| = 1 + 4i

⇒ x + iy + √(x² + y² ) = 1 + 4i

⇒ y = 4 and x + √(x² + y² ) = 1

⇒ x + √(x² + 4² ) = 1

⇒ √(x² + 4² ) = 1 – x

Squaring both sides, we have

⇒ x² + 4² = 1 + x² – 2x

⇒ 2x = -15

or x = -15/2

Example 4:

Solution:

Example 5: Let (α, β) be real, and z be a complex number. If z² + αz + β = 0 has distinct roots on the line Re z = 1, then find the necessary condition.

Solution:

Given Re z = 1

Let z₁ = x + iy and z₂ = x – iy are the roots.

Since Re z = 1, x = 1

Let z² + αz + β = 0 has roots (1 + iy) and (1 – iy)

So,

z₁ z₂ = β

(1 + iy)(1 – iy) = β

1 + y² = β

β > 1

Example 6: If ω(≠ 1) is a cube root of unity, and (1 + ω)⁷ = A + Bω, then find (A, B).

Solution: 

Example 7: A complex number z is said to be unimodular if |z| = 1. Suppose z₁ and z₂ are complex numbers such that

Solution: 

⇒ |z₁| = 2 is a circle of radius 2 and centre at origin.